In this question I asked about temperatures on a planet where a day last 30 days. The data are the same:

  • stellar flux of 1.118
  • albedo between Earth's and Mars's
  • atmospheric composition of 18% oxygen, 13% argon and 69% nitrogen
  • sea-land pressure of 0.87 atmospheres
  • land covered by water is 13%
  • axial tilt and the eccentricity are neglible

If it's useful, this is the map:

enter image description here

How would maximum and minimum temperatures change (between night and day and across latitude) if a day last only 14/15 days? How would this affect the weather?

  • $\begingroup$ How long does a year on the planet last? $\endgroup$ – Nikita Akopjans Jan 22 '16 at 11:06
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    $\begingroup$ if there is no tilt it does not matter (to the weather) $\endgroup$ – King-Ink Jan 22 '16 at 12:15
  • $\begingroup$ So a fortnight is one night $\endgroup$ – TrEs-2b Jan 26 '16 at 21:27
  • $\begingroup$ If "fortnight" means 7 days, yes. $\endgroup$ – Eithne Jan 27 '16 at 13:42

The average equilibrium temperature can be obtained from the Stefan-Boltzmann law, for your data 293.5 K (20 C). Compensating for the Earth-like atmosphere, (+15 K for Earth, closer to +12.5 K for this planet), we have an average temperature of approximately 306 K (33 C). Quite hot, as expected from a higher solar flux, and smaller albedo.

Another useful average we can get from this law is the equatorial average, 311 K without the atmosphere, ~323 K compensated.

Equations for temperature estimations without an atmosphere:

Effective influx: $= solar flux * (1-albedo)$

Global average $= \left(\frac{I_e}{4\sigma}\right)^{\frac{1}{4}}$

Equatorial average $= \left(\frac{I_e}{\pi \sigma}\right)^{\frac{1}{4}}$

Stationary sun-in-zenit average $= \left(\frac{I_e}{\sigma}\right)^{\frac{1}{4}}$

Where $\sigma$ is the Stefan-Boltzmann constant ($σ = 5.67×10^{-8} W m^{-2} K^{-4}$), and $I_e$ is the effective influx.

For a rapidly rotating planet, use the equatorial average for the equator temperature, for a very slowly rotating planet, use the sun-in-zenit equation for the peak temperature. For a case in between those extremes, use something in-between those equations.

Your planet seems to be divided into two regions, a lowland and a highland. We find the highest temperature variations on the equator part of the highland, where a necessary low cloud cover gives huge, dessert like variations, reaching almost 90 C shortly after noon (actually 130 degrees C if we calculate the black-body equilibrium, but we must compensate for atmospheric convection), and less than 0 C degrees (perhaps as low as -15 C) shortly before dawn.

In the lowland, the atmosphere, combined with clouds formed by the lakes ,gives more inertia to the system, thereby limiting the variations. (0 - 50 degrees C).

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  • $\begingroup$ Can you explain where the +90 and -15 came from? I understand how you got the average from the linked law but not where you got the range. $\endgroup$ – Tim B Jan 26 '16 at 9:20
  • $\begingroup$ That is an estimate obtained from the temperature equilibrium for a stationary surface (for the day) and the temperature drop from 14 days of radiation at night. $\endgroup$ – SE - stop firing the good guys Jan 26 '16 at 9:23
  • $\begingroup$ This is a good answer but if there was a bit more detail on how you reached those numbers it would be a great one. For example if it allowed someone coming to the question who had a longer or shorter day than this would they be able to use the information here to produce their own numbers? $\endgroup$ – Tim B Jan 26 '16 at 9:35
  • $\begingroup$ @TimB I added some of the equations, and how to approximate the peak equatorial temperature. Better now? I am new here. $\endgroup$ – SE - stop firing the good guys Jan 26 '16 at 9:48
  • $\begingroup$ Yes, this looks great, most likely this answer will get the bounty :). I'll leave it open for a bit longer though as bounty=more views=more upvotes. Was there any particular method used to compensate for the effects of atmosphere? $\endgroup$ – Tim B Jan 26 '16 at 10:16

You're looking at much the same answer as for your previous question.

A lot of the stability in temperature on Earth is caused by the specific heat capacity of water. It warms up more slowly and cools more slowly than the land. Your planet has relatively very little water, which will cause greater temperature variation to start with. On top of that you've got these long days allowing it to heat up and cool down that much more.

During the day it's going to be hell outside, at night hell will freeze over.

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The Earth maintains nearly-constant average temperature due to its oceans. Not only do the oceans act like a huge heat-sink, storing and releasing energy as needed, they also move energy from the warm equator to the colder poles. Without oceans, the temperature will only be limited by the heat absorbed by the ground - so not a lot. The atmospheric density is slightly lower than Earth's as well, which means even less temperature regulation.

With only 15% of your world being ocean, my guess is that the global climate of the world would somewhat resemble the climate of Mars, though with slightly higher average temperature. Mars' average maximum and minimum temperatures are -7 to -82 C (19 to -116 F) in the winter, and 4 to -73 C (39 to -99 F) in the summer. However, the long days and nights will grow that gap severely.

I don't have any hard numbers, just guesswork, but my guess is that the cold temperatures will reach past -150 C (-238 F), and the hot temperatures will be at or above the boiling point of water. Human life would struggle to survive in domes; outside of a habitat, no human could survive except at dawn and twilight, near the water.

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The daily temperature differences in high and dry places like Colorado's San Luis Valley can swing quite a bit in just 24 hour one summer in the town of Alamosa this happened

August 24, 2002: Record high of 85°, record low of 33° (52° range)

August 25, 2002: Record high of 87°, record low of 30° (57°)

August 26, 2002: Record high of 88°, record low of 31° (57°)

[stats from weather underground]

Back of the envelope gives me the range of boiling point of water to dry ice as possible highs and lows. There also would be a crazy wind blowing from the cold side to the hot as convection currents would be far stronger than earth's. This would damp down the temperature range by a bit by trading it for freezing super hurricanes.

It would be a great place for freeze drying meat.

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  • 2
    $\begingroup$ ºF, I guess, right? $\endgroup$ – JordiVilaplana Jan 22 '16 at 12:52
  • $\begingroup$ @JordiVilaplana Usually I prefer to include both Celsius and Fahrenheit, but in this case it's rather obvious from the record highs that it's Fahrenheit. 88 C would be the equivalent of 190 F. $\endgroup$ – The Anathema Jan 22 '16 at 17:58

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